An electron (mass m) with an initial velocity | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The force on the electron in the electric field is $\overrightarrow{F} = q\overrightarrow{E} = (-e)(-E_{0} \hat{i}) = eE_{0} \hat{i}$.Since the fo

Vedclass · Accepted Answer

$\frac{\lambda_{0}}{\left(1 + \frac{e E_{0} t}{mv_{0}}\right)}$

Vedclass · Answer

(D) The force on the electron in the electric field is $\overrightarrow{F} = q\overrightarrow{E} = (-e)(-E_{0} \hat{i}) = eE_{0} \hat{i}$.Since the force is in the direction of the initial velocity,the electron undergoes uniform acceleration $a = \frac{F}{m} = \frac{eE_{0}}{m}$.The velocity of the electron at time $t$ is $v(t) = v_{0} + at = v_{0} + \frac{eE_{0}t}{m}$.The de Broglie wavelength at time $t$ is given by $\lambda = \frac{h}{mv(t)}$.Substituting the expression for $v(t)$,we get $\lambda = \frac{h}{m(v_{0} + \frac{eE_{0}t}{m})} = \frac{h}{mv_{0} + eE_{0}t}$.Factoring out $mv_{0}$ in the denominator,we get $\lambda = \frac{h}{mv_{0}(1 + \frac{eE_{0}t}{mv_{0}})}$.Since $\lambda_{0} = \frac{h}{mv_{0}}$,we can write $\lambda = \frac{\lambda_{0}}{1 + \frac{eE_{0}t}{mv_{0}}}$.

Similar Questions

Calculate the $(a)$ momentum,and $(b)$ de Broglie wavelength of the electrons accelerated through a potential difference of $56 \; V$.

An electron of mass $m$ with an initial velocity $\vec{V} = V_0 \hat{i} \,(V_0 > 0)$ enters an electric field $\vec{E} = -E_0 \hat{i} \,(E_0 = \text{constant} > 0)$ at $t = 0$. If $\lambda_0$ is its de-Broglie wavelength initially,then its de-Broglie wavelength at time $t$ is:

What is the de Broglie wavelength of an electron accelerated through a potential difference of $ 100 \ V $ (in $\text{Å}$)?

If particles are moving with the same velocity,then the maximum de-Broglie wavelength will be for

The ratio of de-Broglie wavelength of molecules of hydrogen and helium in two jars kept separately at temperatures of $27\,^{\circ}\text{C}$ and $127\,^{\circ}\text{C}$ respectively is

Vedclass Test Series

Exam Paper Generator

Online Exam Module